Answer to Question #255130 in Statistics and Probability for Zah

Question

Glass Houses Pty has found that the cost of fitting new sliding doors is normally distributed, with an average cost per door of R12 750, and a standard deviation of R1 095. A sample of 105 new windows is randomly selected. Assuming normal distribution: What is the probability that the average cost of fitting the doors in

the sample is less than R13 000? Interpret your answer.

Answer 1

Let $\bar{X}=$ the average cost of fitting the doors in the sample: $\bar{X}\sim N(\mu_{\bar{X}}, \sigma_{\bar{X}}^2)$

$$\mu_{\bar{X}}=\mu, \sigma_{\bar{X}}^2=\sigma^2/n$$

Given $\mu=12750, \sigma=1095, n=105.$

$$P(\bar{X}<13000)=P(Z<\dfrac{13000-12750}{1095/\sqrt{105}})$$

$$\approx P(Z<2.3395)\approx0.990345$$

The probability that the average cost of fitting the doors in the sample is less than R13 000 is $0.990345.$